An astronaut is looking down on earth's surface from a space shuttle an altitude of 400 km Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of the size of about :

$$Resolving power = 1.22*\frac{\lambda D}{d}$$ d is the aperture of the instruments D is the distance of satellite from the earth $$Resolving power = \frac{1.22*500*10^{-9}}{5*10^{-3}}$$ $$Resolving power =1.22*\frac{10^{-2}}{10^{-3}}*4$$ =50m